Problem 58 - Chemistry The Central Science

Question

At \(25^{\circ} \mathrm{C}\) the reaction $$\mathrm{CaCrO}_{4}(s) \rightleftharpoons \mathrm{Ca}^{2+}(a q)+\mathrm{CrO}_{4}^{2-}(a q)$$ has an equilibrium constant \(K_{c}=7.1 \times 10^{-4}\). What are the equilibrium concentrations of \(\mathrm{Ca}^{2+}\) and \(\mathrm{CrO}_{4}^{2-}\) in a saturated solution of \(\mathrm{CaCrO}_{4} ?\)

Step-by-Step Solution

Answer

The equilibrium concentrations of both \(Ca^{2+}\) and \(CrO_4^{2-}\) ions in the saturated solution of CaCrO₄ are 0.0266 M.

Step 1: Write the equilibrium expression

First, let's write the equilibrium expression for the given reaction. The equilibrium constant, Kc, is the product of the concentrations of the products divided by the product of the concentrations of the reactants. Based on the balanced chemical equation, the equilibrium expression is: \[K_c = \frac{[\mathrm{Ca}^{2+}][\mathrm{CrO}_{4}^{2-}]}{[\mathrm{CaCrO}_4]}\] Since CaCrO₄ is a solid, its concentration remains unchanged, and we can rewrite the equilibrium expression as: \[K_c = [\mathrm{Ca}^{2+}][\mathrm{CrO}_{4}^{2-}]\]

Step 2: Set up the ICE table

Now, let's set up an ICE (Initial, Change, Equilibrium) table to keep track of the changes in the concentration of the species involved in the reaction. ``` CaCrO₄(s) ⟷ Ca²⁺(aq) + CrO₄²⁻(aq) Initial: - 0 0 Change: - +x +x Equilibrium: - x x ``` In the table, x represents the change in the concentration of each aqueous ion. Note that the concentration of the solid CaCrO₄ remains unchanged.

Step 3: Substitute equilibrium concentrations into the equilibrium expression and solve for x

Now we can substitute the equilibrium concentrations from the ICE table into the equilibrium expression: \[K_c = [\mathrm{Ca}^{2+}][\mathrm{CrO}_{4}^{2-}] = (x)(x) = x^2\] Given the value of Kc: \[7.1 \times 10^{-4} = x^2\] Next, we will find the value of x by taking the square root of both sides: \[x = \sqrt{7.1 \times 10^{-4}}\] Calculating x, we get: \[x = 0.0266\]

Step 4: Determine the equilibrium concentrations of Ca²⁺ and CrO₄²⁻

Now that we have the value of x, we can determine the equilibrium concentrations of Ca²⁺ and CrO₄²⁻. From the ICE table, we know that: \[[\mathrm{Ca}^{2+}] = x = 0.0266\] \[[\mathrm{CrO}_{4}^{2-}] = x = 0.0266\] Therefore, the equilibrium concentrations of Ca²⁺ and CrO₄²⁻ are both 0.0266 M.

Key Concepts

Equilibrium Constant

The equilibrium constant, represented as \(K_c\), is a numerical value that expresses the ratio of the concentration of the products to the reactants at equilibrium. For the given exercise, the equilibrium involves the dissociation of solid calcium chromate (\(\text{CaCrO}_4\)) into its ion components in a saturated solution.
This equilibrium constant bears unique characteristics:

• **Concentration of Solids**: As \(\text{CaCrO}_4\) is a solid, its concentration does not appear in the \(K_c\) expression. This results from the fact that the concentration of a pure solid remains constant during the reaction.
• **Product Only Reaction**: Thus, the equilibrium expression is \(K_c = [\text{Ca}^{2+}][\text{CrO}_4^{2-}]\). This means it is entirely determined by the concentration of the products: calcium ions \([\text{Ca}^{2+}]\) and chromate ions \([\text{CrO}_4^{2-}]\) in the solution.
• **Constant Temperature**: It is crucial to remember that \(K_c\) is dependent on temperature. Here, \(25^{\circ} \text{C}\) is where this equilibrium constant is valid. Any change in temperature will alter the \(K_c\) value.
This simplicity of considering only the soluble ions offers ease when calculating or predicting concentrations at equilibrium.

ICE Table

The ICE table is an invaluable tool when working to understand changes in concentration at equilibrium. ICE is an acronym for Initial, Change, Equilibrium, representing how reactants and products evolve over time during a reaction until equilibrium is reached.

The steps to set up an ICE table include:

• **Initial Concentrations**: Initially, in a saturated solution, the concentrations of the ionic products \(\text{Ca}^{2+}\) and \(\text{CrO}_4^{2-}\) are zero. Solids or pure liquids are not considered in concentration calculations, so \(\text{CaCrO}_4\) is marked with a dash.
• **Change in Concentration**: As the equilibrium establishes, an increase in concentration \((+x)\) happens for \(\text{Ca}^{2+}\) and \(\text{CrO}_4^{2-}\), suggesting that each mole of \(\text{CaCrO}_4\) that dissolves produces equal moles of \(\text{Ca}^{2+}\) and \(\text{CrO}_4^{2-}\).
• **Equilibrium Concentrations**: At equilibrium, both ions have a concentration of \(x\), representing the amount that has dissolved in the solution.
The simplicity of an ICE table allows you to visually track the shift from initial concentration to equilibrium concentration, making these complex systems more digestible.

Solubility Product

The concept of solubility product (\(K_{sp}\)) specifically applies to sparingly soluble salts like \(\text{CaCrO}_4\). It is essentially an equilibrium constant for the dissolution of solid compounds into their constituent ions in a saturated solution.

This specific exercise can be interpreted as an example of finding the solubility of \(\text{CaCrO}_4\) in water.
• **Expression**: For \(\text{CaCrO}_4\), \(K_{sp} = [\text{Ca}^{2+}][\text{CrO}_4^{2-}]\), where the solubility of each ion is captured by \(x\).
• **Dissolution**: The solubility product equation indicates that as \(x\) increases (more dissolution of solid), the concentration of ions also increases in direct proportion, until equilibrium is achieved.
• **Saturated Solution**: At equilibrium, the solution is saturated, meaning no more \(\text{CaCrO}_4\) can dissolve because the maximum ion product consistent with \(K_{sp}\) has been reached.
The solubility product thus serves as a gateway concept that connects the world of solid compounds and their interaction with solvents, highlighting the limits of solubility in precise chemical terms.